2. What is the gravitational force on a 70.0 kg object that is 3.18 x 10 ^6m on the

a load of 800 newton is lifted by an effort of 200 Newton. if the load is placed at a distance of 10 cm from the fulcrum. what w

nataly862011 [7]

Answer:

40 cm

Explanation:

We are given that

Load=800 N

Effort=200 N

Load distance=10 cm

We have to find the effort distance.

We know that

$load\times load\;distance=Effort\times effort\;distance$

Using the formula

$800\times 10=200\times effort\;distance$

Effort distance= $\frac{800\times 10}{200}$

Effort distance= $\frac{8000}{200}$

Effort distance=40 cm

Hence, the effort distance will be 40 cm.

7 0

2 years ago

5.54 Two kilograms of air within a piston–cylinder assembly WP execute a Carnot power cycle with maximum and minimum temper atur

Step2247 [10]

Answer:

A) 60%

B) p2 = 1237.2 kPa

v2 = 0.348 m^3

C) w1-2 = w3-4 = 1615.5 kJ

Q2-3 = 60 kJ

Explanation:

A) calculate thermal efficiency

Л = 1 - $\frac{Tl}{Th}$

where Tl = 300 k

Th = 750 k

hence thermal efficiency ( Л ) = [1 - ( 300 / 750 )] * 100 = 60%

B) calculate the pressure and volume at the beginning of the isothermal expansion

calculate pressure ( P2 ) :

= P3v3 = mRT3 ----- (1)

v3 = 0.4m , mR = 2* 0.287, T3 = 750

hence P3 = 1076.25

next equation to determine P2

Qex = p3v3 ln( p2/p3 )

60 = 1076.25 * 0.4 ln(p2/p3)

hence ; P2 = 1237.2 kpa

calculate volume ( V2 )

p2v2 = p3v3

v2 = p3v3 / p2

= (1076.25 * 0.4 ) / 1237.2

= 0.348 m^3

C) calculate the work and heat transfer for each four processes

work :

W1-2 = mCv( T2 - T1 )

= 2*0.718 ( 750 - 300 ) = 1615.5 kJ

W3-4 = 1615.5 kJ

heat transfer

Q2-3 = W2-3 = 60KJ

Q3-4 = 0

D ) sketch of the cycle on p-V coordinates

attached below

6 0

2 years ago

Three students have been studying relative motion and decide to do an experiment to demonstrate their knowledge. The experiment

miskamm [114]

Answer with Explanation:

We are given that

Constant speed of Jane=12.6 m/s

a.When Fred can throw the ball 30 m/s

We have to find the angle relative to the horizontal when he throw the ball in order for Sue to see the ball travel vertically upward.

Let $\theta$ be the angle .

Therefore,

$30 cos\theta=12.6$

$cos\theta=\frac{12.6}{30}=0.42$

$\theta=cos^{-1}(0.42)=65.165^{\circ}$

b.We have to find the height to which ball reach.

$v^2-v^2_0=2aS$

$S=\frac{v^2-v^2_0}{2a}=\frac{0-(30 sin65.165)^2}{2(-9.81)}$

$S=37.78 m$

7 0

3 years ago

Question 20 Unsaved The graph below represents changes in water. What process is occurring during section "B" of the graph? Ques

FinnZ [79.3K]

melting ... happpens at constant temp

4 0

2 years ago

Read 2 more answers

A wire is wrapped around a piece of iron, and then electricity is run through the wire. What happens to the iron?

katrin [286]

Search Results<span>By simply wrapping wire that has an electrical current running through it around a nail, you can make an electromagnet. When the electric current moves through a wire, it makes a magnetic field. ... You can make a temporary magnet by stroking apiece of iron or steel (such as a needle) along with a permanent magnet.

Hope This Helps!</span>

7 0

2 years ago